Set up arrangment of three point charges : `q, + 2q and x q` separated by equal finite distances so that electric potential energy of the system is zero. What is x ?

Options

A. `-2/3`
B. `-1/3`
C. `2/3`
D. `3/2`

Answer

To solve the problem of finding the value of \( x \) such that the electric potential energy of the system of three point charges \( q \), \( +2q \), and \( xq \) is zero, we can follow these steps:

### Step 1: Arrangement of Charges
We will arrange the three charges \( q \), \( +2q \), and \( xq \) at the vertices of an equilateral triangle. Let’s denote:
- Charge \( q \) at vertex A,
- Charge \( +2q \) at vertex B,
- Charge \( xq \) at vertex C.

### Step 2: Expression for Potential Energy
The potential energy \( U \) of a system of point charges is given by the sum of the potential energies between each pair of charges. The potential energy \( U \) can be expressed as:
\[ U = k \left( \frac{q \cdot 2q}{r} + \frac{q \cdot xq}{r} + \frac{2q \cdot xq}{r} \right) \]
where \( k = \frac{1}{4\pi \epsilon_0} \) and \( r \) is the distance between the charges.

### Step 3: Simplifying the Expression
Substituting the values into the equation:
\[ U = k \left( \frac{2q^2}{r} + \frac{xq^2}{r} + \frac{2xq^2}{r} \right) \]
This simplifies to:
\[ U = k \left( \frac{2q^2 + xq^2 + 2xq^2}{r} \right) \]
\[ U = k \left( \frac{2q^2 + 3xq^2}{r} \right) \]
Factoring out \( q^2 \):
\[ U = k \left( \frac{q^2(2 + 3x)}{r} \right) \]

### Step 4: Setting Potential Energy to Zero
For the potential energy \( U \) to be zero:
\[ \frac{q^2(2 + 3x)}{r} = 0 \]
Since \( q^2 \) and \( r \) are not zero, we can set the term in parentheses to zero:
\[ 2 + 3x = 0 \]

### Step 5: Solving for \( x \)
Now, solving for \( x \):
\[ 3x = -2 \]
\[ x = -\frac{2}{3} \]

### Conclusion
Thus, the value of \( x \) that makes the electric potential energy of the system zero is:
\[ \boxed{-\frac{2}{3}} \]